Uniform convergence of $\frac{y/(2N)}{\sin(y/(2N))}$ towards 1

Question

I can't come up with a proof, why $f_N(y) := \frac{\frac{y}{2N}}{\sin\left(\frac{y}{2N}\right)}$ converges uniformly against $1$ for $y\in(0,\pi),\ N\to\infty$.

I would be thankful for any advice.

Answer 1

A solution using Dini's theorem

The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,\pi]$.

For $n \ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x \in [0, \pi]$ the sequence $(1/f_n(x))$ is increasing.